The graph shows ln left| frac{dN(t)}{dt} righ | vedclass

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(B) The activity of a radioactive sample is given by $A = \left| \frac{dN(t)}{dt} \right| = \lambda N_0 e^{-\lambda t}$.Taking the natural logarithm o

Vedclass · Accepted Answer

$8$

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(B) The activity of a radioactive sample is given by $A = \left| \frac{dN(t)}{dt} ight| = \lambda N_0 e^{-\lambda t}$.Taking the natural logarithm on both sides,we get $\ln A = \ln(\lambda N_0) - \lambda t$.This is a linear equation of the form $y = mx + c$,where the slope $m = -\lambda$.From the given graph,the slope is $	ext{slope} = \frac{3 - 4}{6 - 4} = \frac{-1}{2}$.Thus,$-\lambda = -\frac{1}{2}$,which gives the decay constant $\lambda = 0.5 	ext{ year}^{-1}$.The half-life is $T_{1/2} = \frac{\ln 2}{\lambda} = \frac{0.693}{0.5} = 1.386 	ext{ years}$.The number of nuclei remaining after time $t$ is $N(t) = N_0 e^{-\lambda t} = N_0 \left( \frac{1}{2} ight)^{t/T_{1/2}}$.Given $t = 4.16 	ext{ years}$,the number of half-lives elapsed is $n = \frac{4.16}{1.386} \approx 3$.Therefore,the remaining fraction is $\frac{N(t)}{N_0} = \left( \frac{1}{2} ight)^3 = \frac{1}{8}$.Since the number of nuclei decreases by a factor of $p$,we have $\frac{N(t)}{N_0} = \frac{1}{p} = \frac{1}{8}$,which implies $p = 8$.

The graph shows $\ln \left| \frac{dN(t)}{dt} \right|$ versus $t$ for a radioactive sample. If the number of radioactive nuclei decreases by a factor of $p$ after $4.16$ years,then $p = $.....

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